a train goes into a tunnel at 20m/s and emerges from itat 55m/s. the tunnel is 1500m long.assuming constant acceleration the time for which the train moves in tunnel is
a=(V^2-Vo^2)/2d
a=(55^2-20^2)/3000=0.875m/s^2
V = Vo + a*t
t = (V-Vo)/a = (55-20)/0.875 = 40 s.
To find the time for which the train moves in the tunnel, we can use the equation:
v = u + at
Where:
v = final velocity (55 m/s)
u = initial velocity (20 m/s)
a = acceleration (constant acceleration)
t = time
Since the train is moving with constant acceleration, we can use the formula:
v^2 = u^2 + 2as
Where:
s = distance (1500 m)
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
a = (55^2 - 20^2) / (2 * 1500)
a = (3025 - 400) / 3000
a = 2625 / 3000
a = 0.875 m/s^2
Now we can substitute the values of u, v, and a into the first equation to solve for time (t):
55 = 20 + (0.875)t
Subtracting 20 from both sides:
35 = 0.875t
Dividing both sides by 0.875:
t = 35 / 0.875
t = 40 seconds
Therefore, the train moves in the tunnel for 40 seconds.
To find the time for which the train moves in the tunnel, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) of the train is 20 m/s, the final velocity (v) is 55 m/s, and the acceleration (a) is constant. We need to find the time (t).
First, let's find the acceleration using the equation:
v = u + at
55 = 20 + a * t
Rearrange the equation:
a * t = 55 - 20
a * t = 35
Next, we need the value of acceleration (a). We can use the equation of motion again:
a = (v - u) / t
Substituting the known values:
35 = (55 - 20) / t
Let's solve for t:
35t = 55 - 20
35t = 35
t = 1
The time for which the train moves in the tunnel is 1 second.